, 2012

Let $I$ be the center of the inscribed circle of the triangle $ABC$, and let $K$ and $L$ be the intersection points of the line $DE$ with the lines $AB$ and $AC$, respectively. Because the bisector of an angle of a triangle goes through the center of the inscribed circle and the midpoint of the opposite arc, the points $B, I$ and $E$ lie on a common line (the bisector of the angle at $B$). Hence $\angle IBK = \angle CBI = \angle CDE$, and the points $B, I, K$ and $D$ are concyclic. It may be shown similarly that the points $C, I$ and $D$ lie on a common line (the bisector of the angle at $C$) and that the points $C, E, L$ and $I$ are concyclic.

From this we conclude $\angle LKA = \angle DKB = \angle DIB = \angle CIE = \angle CLE = \angle ALK$. The triangle $KLA$ is thus isosceles with the top angle at $A$. Because $AF$ and $AG$ are tangent segments, the triangle $FGA$ is also isosceles with the top angle at $A$. The

lines $DE$ and $FG$ are thus parallel. According to the assumptions of the problem, the triangle $EDX$ is isosceles with the top angle at $X$, hence the triangle $GFX$ is also isosceles with the top angle at $X$. The quadrilateral $AFXG$ is thus a deltoid, and the line $AX$ is perpendicular to $GF$ and, consequently, it is also perpendicular to $DE$. From this we conclude that $ADXE$ is a deltoid, hence $ADE$ is an isosceles triangle with the top angle at $A$. The angles $\angle ACD$ and $\angle EBA$ are equal because they are inscribed angles over chords of equal length. The triangle $ABC$ is thus an isosceles triangle with the top angle at $A$.